Solution manual fundamentals of physics extended, 8th editionch19

101 13 0
  • Loading ...
1/101 trang
Tải xuống

Thông tin tài liệu

Ngày đăng: 13/09/2018, 13:37

1 Each atom has a mass of m = M/NA, where M is the molar mass and NA is the Avogadro constant The molar mass of arsenic is 74.9 g/mol or 74.9 × 10–3 kg/mol 7.50 × 1024 arsenic atoms have a total mass of (7.50 × 1024) (74.9 × 10–3 kg/mol)/(6.02 × 1023 mol–1) = 0.933 kg (a) Eq 19-3 yields n = Msam/M = 2.5/197 = 0.0127 mol (b) The number of atoms is found from Eq 19-2: N = nNA = (0.0127)(6.02 × 1023) = 7.64 × 1021 (a) We solve the ideal gas law pV = nRT for n: ( ) (100 Pa ) 1.0 ×10−6 m3 pV n= = = 5.47 × 10−8 mol RT ( 8.31J/mol ⋅ K )( 220 K ) (b) Using Eq 19-2, the number of molecules N is ( N = nN A = 5.47 × 10−6 mol ) ( 6.02 × 10 23 ) mol−1 = 3.29 × 1016 molecules With V = 1.0 × 10–6 m3, p = 1.01 × 10–13 Pa, and T = 293 K, the ideal gas law gives ( )( ) 1.01 × 10−13 Pa 1.0 × 10−6 m3 pV = = 4.1 × 10 −23 mole n= RT 8.31 J/mol ⋅ K 293 K ( )( ) Consequently, Eq 19-2 yields N = nNA = 25 molecules We can express this as a ratio (with V now written as cm3) N/V = 25 molecules/cm3 (a) In solving pV = nRT for n, we first convert the temperature to the Kelvin scale: T = (40.0 + 273.15) K = 313.15 K And we convert the volume to SI units: 1000 cm3 = 1000 × 10–6 m3 Now, according to the ideal gas law, ( )( ) 1.01 × 105 Pa 1000 × 10−6 m3 pV n= = = 3.88 × 10−2 mol RT (8.31J/mol ⋅ K )( 313.15 K ) (b) The ideal gas law pV = nRT leads to ( ( )( ) 1.06 × 105 Pa 1500 × 10−6 m3 pV T= = = 493K nR 3.88 × 10−2 mol ( 8.31J/mol ⋅ K ) ) We note that the final temperature may be expressed in degrees Celsius as 220°C Since (standard) air pressure is 101 kPa, then the initial (absolute) pressure of the air is pi = 266 kPa Setting up the gas law in ratio form (where ni = nf and thus cancels out — see Sample Problem 19-1), we have pfVf piVi = Đ 1.64 ì 102 m3 · § 300 K · Ÿ p f = ( 266 kPa ) ă á ă Ti â 1.67 ì 10 m â 273K Tf which yields pf = 287 kPa Expressed as a gauge pressure, we subtract 101 kPa and obtain 186 kPa (a) With T = 283 K, we obtain (100 ´ 103 Pa )(2.50 m3 ) pV n= = = 106 mol RT (8.31J/mol ×K )(283 K ) (b) We can use the answer to part (a) with the new values of pressure and temperature, and solve the ideal gas law for the new volume, or we could set up the gas law in ratio form as in Sample Problem 19-1 (where ni = nf and thus cancels out): pfVf piVi = Tf § 100 kPa · § 303K ã V f = 2.50 m3 ă ă Ti â 300 kPa â 283K ( which yields a final volume of Vf = 0.892 m3 ) (a) Eq 19-45 (which gives 0) implies Q = W Then Eq 19-14, with T = (273 + 30.0)K leads to gives Q = –3.14 × 103 J, or | Q | = 3.14 × 103 J (b) That negative sign in the result of part (a) implies the transfer of heat is from the gas The pressure p1 due to the first gas is p1 = n1RT/V, and the pressure p2 due to the second gas is p2 = n2RT/V So the total pressure on the container wall is p = p1 + p2 = n1 RT n2 RT RT + = ( n1 + n2 ) V V V The fraction of P due to the second gas is then p2 n2 RT / V n2 0.5 = = = = 0.2 p ( n1 + n2 )( RT / V ) n1 + n2 + 0.5 10 Using Eq 19-14, we note that since it is an isothermal process (involving an ideal gas) then Q = W = nRT ln(Vf /Vi) applies at any point on the graph An easy one to read is Q = 1000 J and Vf = 0.30 m3, and we can also infer from the graph that Vi = 0.20 m3 We are told that n = 0.825 mol, so the above relation immediately yields T = 360 K 80 We label the various states of the ideal gas as follows: it starts expanding adiabatically from state until it reaches state 2, with V2 = m3; then continues on to state isothermally, with V3 = 10 m3; and eventually getting compressed adiabatically to reach state 4, the final state For the adiabatic process → p1V1γ = p2V2γ , for the isothermal process → p2V2 = p3V3, and finally for the adiabatic process → p3V3γ = p4V4γ These equations yield γ γ γ γ §V ·§ V · §V · § V · § V ·§ V ã p4 = p3 ă = p2 ¨ ¸ ¨ ¸ = p1 ¨ ă ă â V4 ¹ © V2 ¹ © V3 ¹ © V4 ¹ © V3 ¹ © V4 ¹ We substitute this expression for p4 into the equation p1V1 = p4V4 (since T1 = T4) to obtain V1V3 = V2V4 Solving for V4 we obtain ( )( ) 2.0 m 10 m V1V3 V4 = = = 5.0 m3 V2 4.0 m3 81 We write T = 273 K and use Eq 19-14: § 16.8 · W = (1.00 mol ) ( 8.31 J/mol ⋅ K ) ( 273 K ) ln ă â 22.4 which yields W = –653 J Recalling the sign conventions for work stated in Chapter 18, this means an external agent does 653 J of work on the ideal gas during this process 82 (a) We use pV = nRT The volume of the tank is nRT V= = p ( 300g 17 g mol ) (8.31 J/mol ⋅ K )( 350 K ) = 3.8 ×10 1.35 ×106 Pa −2 m3 = 38 L (b) The number of moles of the remaining gas is n′ = −2 p′V ( 8.7 × 10 Pa )( 3.8 ×10 m ) = = 13.5 mol RT ′ ( 8.31 J/mol ⋅ K )( 293K ) The mass of the gas that leaked out is then ∆m = 300 g – (13.5 mol)(17 g/mol) = 71 g 83 From Table 19-3, CV = 32 R = 12.5 J/mol ⋅ K for a monatomic gas such as helium To obtain the desired result cV we need to effectively “convert” mol → kg, which can be done using the molar mass M expressed in kilograms per mole Although we could look up M for helium in Table 19-1 or Appendix F, the problem gives us m so that we can use Eq 19-4 to find M That is, M = mN A = ( 6.66 ×10 −27 kg )( 6.02 ×1023 / mol ) = 4.01× 10−3 Therefore, cV = CV/M = 3.11 × 103 J/kg·K kg mol 84 (a) When n = 1, V = Vm = RT/p, where Vm is the molar volume of the gas So Vm = RT ( 8.31J/mol ⋅ K )( 273.15 K ) = = 22.5 L 1.01×105 Pa p (b) We use vrms = 3RT / M The ratio is given by vrms, He vrms, Ne = M Ne 20 g = = 2.25 M He 4.0 g (c) We use λ Ηe = ( 2πd N V ) −1 , where the number of particles per unit volume is given by N/V = NAn/V = NAp/RT= p/kT So λ He = = 2πd ( p / kT ) = kT 2πd p (1.38×10−23 J/K ) ( 273.15 K ) = 0.840 m 1.414 (1ì1010 m ) (1.01×105 Pa ) (d) λNe = λHe = 0.840 µm 85 For convenience, the “int” subscript for the internal energy will be omitted in this solution Recalling Eq 19-28, we note that ¦E = cycle ∆E A→ B + ∆EB→C + ∆EC → D + ∆ED→ E + ∆EE → A = Since a gas is involved (assumed to be ideal), then the internal energy does not change when the temperature does not change, so ∆E A→ B = ∆ED→ E = Now, with ∆EE→A = 8.0 J given in the problem statement, we have ∆EB →C + ∆EC → D + 8.0 J = In an adiabatic process, ∆E = –W, which leads to −5.0 J + ∆EC → D + 8.0 J = 0, and we obtain ∆EC→D = –3.0 J 86 We solve 3RT 3R(293K) = M helium M hydrogen for T With the molar masses found in Table 19-1, we obtain Đ 4.0 ã T = (293K) ă = 580 K â 2.02 which is equivalent to 307°C 87 It is straightforward to show, from Eq 19-11, that for any process that is depicted as a straight line on the pV diagram, the work is § pi + p f ã Wstraight = ă V © ¹ which includes, as special cases, W = p∆V for constant-pressure processes and W = for constant-volume processes Also, from the ideal gas law in ratio form (see Sample Problem 1), we find the final temperature: § p ãĐV ã T2 = T1 ă ă = 4T1 â p1 â V1 ¹ (a) With ∆V = V2 – V1 = 2V1 – V1 = V1 and p1 + p2 = p1 + 2p1 = 3p1, we obtain W= 3 W = = 1.5 ( p1 V1 ) = nRT1 Ÿ nRT1 2 where the ideal gas law is used in that final step (b) With ∆T = T2 – T1 = 4T1 – T1 = 3T1 and CV = 32 R , we find ∆Eint 9 Đ3 ã Eint = n ă R ( 3T1 ) = nRT1 = = 4.5 â2 nRT1 (c) The energy added as heat is Q = ∆Eint + W = 6nRT1, or Q / nRT1 = (d) The molar specific heat for this process may be defined by C= Q 6nRT1 C = = 2R Ÿ = n∆T n (3T1 ) R 88 The gas law in ratio form (see Sample Problem 19-1) leads to ĐV ã p2 = p1 ă â V2 Đ 4.00 m3 ã Đ T2 ã 5.67 Pa = ( ) ă ăT â 7.00 m â 1ạ Đ 313 K ã ă = 4.67 Pa â 217 K 89 It is recommended to look over §19-7 before doing this problem (a) We normalize the distribution function as follows: ³ vo P ( v ) dv = Ÿ C = vo3 (b) The average speed is ³ vo vP ( v ) dv = vo Đ 3v ã v ă dv = vo â vo (c) The rms speed is the square root of ³ vo vo § 3v · ( ) v P v dv = v ă dv = vo2 â vo Therefore, vrms = 5vo ≈ 0.775vo 90 (a) From Table 19-3, CV = 52 R and C p = 72 R Thus, Eq 19-46 yields §7 · Q = nC p ∆T = ( 3.00 ) ă ( 8.31) ( 40.0 ) = 3.49 ì 103 J â2 (b) Eq 19-45 leads to Đ5 ã Eint = nCV T = ( 3.00 ) ă ( 8.31) ( 40.0 ) = 2.49 ì 103 J â2 (c) From either W = Q ∆Eint or W = p∆T = nR∆T, we find W = 997 J (d) Eq 19-24 is written in more convenient form (for this problem) in Eq 19-38 Thus, we obtain §3 · ∆K trans = ∆ ( NK avg ) = n ă R T 1.50 ì 103 J â2 91 (a) The temperature is 10.0C → T = 283 K Then, with n = 3.50 mol and Vf/V0 = 3/4, we use Eq 19-14: § Vf ã W = nRT ln ă = 2.37 kJ â V0 (b) The internal energy change ∆Eint vanishes (for an ideal gas) when ∆T = so that the First Law of Thermodynamics leads to Q = W = –2.37 kJ The negative value implies that the heat transfer is from the sample to its environment 92 (a) Since n/V = p/RT, the number of molecules per unit volume is N nN A § p = = NA ă V V â RT 1.01 ì 105 Pa molecules · 23 (6.02 × 10 ) = 2.5 ì 1025 J m3 (8.31 molK )( 293K ) (b) Three-fourths of the 2.5 ì 1025 value found in part (a) are nitrogen molecules with M = 28.0 g/mol (using Table 19-1), and one-fourth of that value are oxygen molecules with M = 32.0 g/mol Consequently, we generalize the Msam = NM/NA expression for these two species of molecules and write 28.0 32.0 (2.5 × 1025 ) + (2.5 × 1025 ) = 1.2 × 103 g 23 6.02 × 10 6.02 × 1023 93 (a) The work done in a constant-pressure process is W = p∆V Therefore, W = ( 25 N/m ) (1.8 m3 − 3.0 m3 ) = − 30 J The sign conventions discussed in the textbook for Q indicate that we should write –75 J for the energy which leaves the system in the form of heat Therefore, the first law of thermodynamics leads to ∆Eint = Q − W = (−75 J) − (−30 J) = −45 J (b) Since the pressure is constant (and the number of moles is presumed constant), the ideal gas law in ratio form (see Sample Problem 19-1) leads to ĐV ã Đ 1.8m ã T2 = T1 ă = (300 K) ă = 1.8 ì102 K â 3.0 m â V1 It should be noted that this is consistent with the gas being monatomic (that is, if one assumes CV = 32 R and uses Eq 19-45, one arrives at this same value for the final temperature) 94 Since no heat is transferred in an adiabatic process, then § 3· Qtotal = Qisotherm = Wisotherm = nRT ln ă â 12 ¹ where the First Law of Thermodynamics (with ∆Eint = during the isothermal process) and Eq 19-14 have been used With n = 2.0 mol and T = 300 K, we obtain Q = –6912 J ≈ –6.9 kJ ... number of moles of gas in the bubble is n = p1V1/RT1 = (p0 + ρgd)V1/RT1, where V1 is the volume of the bubble at the bottom of the lake and T1 is the temperature there At the surface of the lake... is the heat of vaporization and N is the number of molecules per gram The molar mass of atomic hydrogen is g/mol and the molar mass of atomic oxygen is 16 g/mol so the molar mass of H2O is (1.0... from Vf to Vi Now Vf = piVi/pf, so æ p V ửữ W = p f ỗỗỗVi - i i ÷ = ( p f - pi )Vi = (1.013´ 105 Pa - 2.04´ 105 Pa )(0.14 m3 ) ÷ ữ ỗố pf ứ = - 1.44 104 J The total work done by the gas over the entire
- Xem thêm -

Xem thêm: Solution manual fundamentals of physics extended, 8th editionch19 , Solution manual fundamentals of physics extended, 8th editionch19

Từ khóa liên quan

Gợi ý tài liệu liên quan cho bạn

Nhận lời giải ngay chưa đến 10 phút Đăng bài tập ngay